Estimation of power of number by its factorial

Question

Let $a \in R$, such as $a>1$. How to find smallest natural number $N$, such that $a^N < N!$?

Answer 1

by taking the logarithm on both sides we obtain $$N\ln(a)<\ln(N!)$$ from here we get $$\ln(a)<\frac{\ln(N!)}{N}$$ thus we get $$a<e^{\frac{\ln(N!)}{N}}$$

Answer 2

In fact @Robert Israel has found a formula to the sequence.

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A small check of the smallest $N$ for $a =1,2,\cdots 10$ gives us the sequence $A065027$ in the OEIS giving us the smallest $N >0$ such that $a^N <N! $.

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He writes that it appears that $L(n) < a(n) - n e + \log{\sqrt{2 \pi n)}} < \frac {1}{2} $, where $L(n) = -\frac {1}{2} + o(1)$ , and $L(n) > -0.53$ for all $n$.

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Hope it helps.